Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $y = \dfrac{k + 5}{-6k - 6} \times \dfrac{k^2 + 9k + 8}{k + 5} $
Explanation: First factor the quadratic. $y = \dfrac{k + 5}{-6k - 6} \times \dfrac{(k + 1)(k + 8)}{k + 5} $ Then factor out any other terms. $y = \dfrac{k + 5}{-6(k + 1)} \times \dfrac{(k + 1)(k + 8)}{k + 5} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (k + 5) \times (k + 1)(k + 8) } { -6(k + 1) \times (k + 5) } $ $y = \dfrac{ (k + 5)(k + 1)(k + 8)}{ -6(k + 1)(k + 5)} $ Notice that $(k + 5)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ \cancel{(k + 5)}(k + 1)(k + 8)}{ -6\cancel{(k + 1)}(k + 5)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $y = \dfrac{ \cancel{(k + 5)}\cancel{(k + 1)}(k + 8)}{ -6\cancel{(k + 1)}\cancel{(k + 5)}} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $y = \dfrac{k + 8}{-6} $ $y = \dfrac{-(k + 8)}{6} ; \space k \neq -1 ; \space k \neq -5 $